partial fraction decomposition, decompose denominator I would like to compute the following integral 
$$\displaystyle \int_{\gamma} \frac{z-1}{z(z-i)(z-3i)}dz$$ where $\gamma$ is the curve $|z-i|=1/2$ 

My question: How can i decompose it to partial fraction, i tried to multiply the nominator with $(z)(z-i)(z-3i)$ so $$A\cdot(z-i)(z-3i)+B\cdot z(z-3i)+C\cdot z(z-i)=z-1$$ and then compare the variable ,but really takes a lot of time,  is there another way to find $A$,$B$ and $C$? such thatwe have the equality:

$$\displaystyle \int_{\gamma} \frac{z-1}{z(z-i)(z-3i)}dz=\displaystyle \int_{\gamma}\left( \frac{A}{z}dz+ \frac{B}{z-i}dz+ \frac{C}{z-3i}\right)dz$$ 
I got by comparing the following:
$$\displaystyle \int_{\gamma}\left( \frac{1}{3z}dz+ \frac{i-1}{2(z-i)}dz+ \frac{3i-1}{6(z-3i)}\right)dz$$
Thanks!!
 A: You can use the residue method
$$ A = \operatorname{Res}\big(f(z),0\big) = \lim_{z\to 0} z\ f(z) = \frac{1}{3} $$
and so on
A: $$\frac{z-1}{z(z-i)(z-3i)}=\frac{A}{z}+\frac{B}{z-i}+\frac{C}{z-3i}$$
To get $A$ multiply both sides by $z$ and then take the limit as $z\to 0$ in both sides:
Let $f(z)=\displaystyle{\frac{z-1}{z(z-i)(z-3i)}}$. Then, par example
$$ z\cdot f(z) = A +z\cdot\frac{B}{z-i}+z\cdot\frac{C}{z-3i} $$
$$ (z-i)\cdot f(z) = (z-i)\cdot\frac{A}{z}+B+(z-i)\cdot\frac{C}{z-3i} $$
$$ (z-3i)\cdot f(z) = (z-3i)\cdot\frac{A}{z} +(z-3i)\cdot\frac{B}{z-i}+C $$
Taking the corresponding limit in each case gives you:
$$A=\underset{z\to 0}{\lim}{\ z\cdot f(z)}=\frac{1}{3}$$
$$B=\underset{z\to i}{\lim}{\ (z-i)\cdot f(z)}=\frac{i-1}{i(-2i)}=\frac{i-1}{2}$$
$$C=\underset{z\to 3i}{\lim}{\ (z-3i)\cdot f(z)}=\frac{3i-1}{3i(2i)}=\frac{3i-1}{-6}=\frac{1-3i}{6}$$
Also, just starting eith your equation:
$$A\,(z-i)(z-3i)+B\,z(z-3i)+C\,z(z-i)=z-1$$
You can:
1) Evaluate at $z=0$ both sides to isolate after $A$
2) Evaluate at $z=i$ both sides to isolate after $B$
3) Evaluate at $z=3i$ both sides to isolate after $C$
